This fact is actually true even if $f$ is not continuous but attains its maximum and minimum in every closed interval (for example, $f(x)=x+\operatorname{sgn}x$).
First, notice that $f$ is injective: If we could find $x_1<x_2$ such that $f(x_1)=f(x_2)$, then $f$ would necessarily attain both its maximum and minimum in the open interval $(x_1,x_2)$.
Let $a<b<c$ in $\mathbb{R}$ be arbitrary. Let $M=\max f[a,c]$ and $m=\min f[a,c]$. Since $f$ does not attains it maximum nor its minimum in the open interval $(a,c)$, then either $(f(a),f(c))=(m,M)$ or $(f(a),f(c))=(M,m)$.
Now, let's show that the following are equivalent:
(i) $f(a)<f(b)$.
(ii) $f(b)<f(c)$.
(iii) $f(a)<f(c)$.
Proof:
(i)$\Rightarrow$(ii). Then $f(a)<f(b)\leq M$, so $f(a)\neq M$, thus $f(a)=m$ and $f(c)=M>f(b)$.
(ii)$\Rightarrow$(iii). Then $m\leq f(b)<f(c)$, so $f(c)\neq m$, thus $f(c)=M$ and $f(a)=m<M=f(c)$.
(iii)$\Rightarrow$(i). In this case, necessarily $f(a)=m$ and $f(c)=M$, so in particular $f(a)=m\leq f(b)$, and, since $f$ is injective, $f(a)\neq f(b)$, so $f(a)<f(b)$.
Finally, by changing $f$ by $-f$ is necessary, we may suppose that $f(0)<f(1)$. Let's show that $f$ is (strictly) increasing: Let $x<y$ in $\mathbb{R}$ be arbitrary. Let $K=|x|+|y|+1$. Notice that $-K<x<y<K$.
Using $a=0$, $b=1$, $c=K$ and $(i)\Rightarrow(iii)$ as above, we obtain $f(0)<f(K)$.
Using $a=-K$, $b=0$, $c=K$ and $(ii)\Rightarrow(iii)$ as above, we obtain $f(-K)<f(K)$.
Using $a=-K$, $b=x$, $c=K$ and $(iii)\Rightarrow (i)$ as above, we obtain $f(-K)<f(x)$.
Finally, using $a=-K$, $b=x$, $c=y$ and $(i)\Rightarrow(ii)$ as above, we obtain $f(x)<f(y)$.
Therefore, $f$ is strictly increasing, hence monotone.
Here is a simple solution.
Let $\{x_n\}\in [0,1]$ be a minimizing sequence, that is, such that
$$\lim_{n\to \infty} f(x_n)= f^*= \inf_{x\in [0,1]}f(x).$$ Such a sequence always exists by the definition of infimum. Since $[0,1]$ is compact, the sequence $\{x_n\}$ has a convergent subsequence. Without loss of generality assume that $x_n\to \bar{x}\in [0,1].$ Then
$$f(\bar{x})\leq \liminf_{n\to \infty}f(x_n)= \lim_{n\to \infty} f(x_n)= f^*.$$ Since $\bar{x} \in [0,1]$ and by the definition of $f^*,$ we can only have
$$f(\bar{x})=f^*,$$ as desired.
Best Answer
$f$ is (inferior) bounded by some $M>0$ so for $\epsilon_0>0$ small enough $M/\epsilon > f(1), 0<\epsilon \le \epsilon_0$; this means that the infimum of $f(x)/x$ on $(0,1]$ happens on $[\epsilon_0,1]$ but there $f(x)/x$ is continuos so the infimum is a minimum and is attained. Done!